How many practice questions are available for Triangles?

SparkEd Math offers 60+ practice questions for Triangles across 3 difficulty levels (Easy, Medium, Hard) plus a timed test mode. Each question comes with visual step-by-step solutions and AI coaching hints.

Is Triangles content free on SparkEd Math?

Yes! All Triangles content on SparkEd Math is 100% free, including practice questions, visual solutions, lessons, and AI coaching. No credit card required.

How does SparkEd Math teach Triangles differently from textbooks?

SparkEd Math uses visual step-by-step solutions for every Triangles problem, making complex concepts easy to understand. The platform includes 3 difficulty levels for progressive learning, a Super Power Help AI coach for instant doubt-clearing, and interactive practice rather than passive reading.

Is SparkEd's Triangles content aligned with the Olympiad syllabus?

Yes, all Triangles questions and lessons on SparkEd Math are fully aligned with the Mathematical Olympiad Preparation (Olympiad) syllabus for Class 10. The content follows the official curriculum structure and exam pattern.

Triangles — Class 10 Olympiad (2025-26)

About Triangles — Class 10 Olympiad

Apply similarity criteria (AA, SAS, SSS), BPT, and Pythagoras theorem in advanced geometry problems. This topic is part of the Olympiad Class 10 mathematics syllabus (chapter: Module 6). On this page you can practice 58 questions across three difficulty levels — 20 easy, 18 medium, and 20 hard — each with a visual step-by-step solution, plus a timed 34-question mastery test with a live leaderboard. Worked examples and sample questions from the topic are below.

Triangles — solved examples for Class 10 Olympiad

Example 1easy

In triangle PQR, S is a point on PQ and T is a point on PR such that ∠PST = ∠PRQ. If PS = 6 cm, PQ = 12 cm, and PR = 10 cm, what is the length of PT?

A)5 cm
B)6 cm
C)7 cm
D)7.2 cm

Step-by-step solution

In ΔPST and ΔPQR, we have ∠P as common to both triangles. We are given that ∠PST = ∠PRQ.
By the Angle-Angle (AA) similarity criterion, ΔPST ~ ΔPQR.
The ratio of corresponding sides must be equal: PS/PQ = PT/PR.
Substitute the given values: 6/12 = PT/10. Simplifying, 1/2 = PT/10. Therefore, PT = 5 cm.

Answer: 5 cm

Example 2medium

In triangle ABC, points D and E are on sides AB and AC respectively, such that DE || BC. Point F is on BC such that EF || AB. If the ratio AD/DB = 2/3, what is the ratio of the area of triangle ADE to the area of quadrilateral BDFE?

A)1/3
B)2/5
C)3/8
D)4/9

Step-by-step solution

Given DE || BC and AD/DB = 2/3. By BPT, AE/EC = AD/DB = 2/3. So, AD/AB = 2/5 and AE/AC = 2/5.
Since DE || BC, triangle ADE is similar to triangle ABC. The ratio of their areas is (AD/AB)².
$Area(ΔADE) / Area(ΔABC) = (AD/AB)² = (2/5)² = 4/25.$
Given EF || AB. This implies triangle CEF is similar to triangle CAB. The ratio of similarity CE/CA. From AE/EC = 2/3, we have AC = AE + EC = (2/3)EC + EC = (5/3)EC, so EC/AC = 3/5. Thus, the ratio of their areas is (EC/AC)².
$Area(ΔCEF) / Area(ΔCAB) = (EC/AC)² = (3/5)² = 9/25.$
Area(BDFE) = Area(ΔABC) - Area(ΔADE) - Area(ΔCEF). So, Area(BDFE) = Area(ΔABC) × (1 - 4/25 - 9/25) = Area(ΔABC) × (12/25). The required ratio is Area(ΔADE) / Area(BDFE) = (4/25) / (12/25) = 4/12 = 1/3.

Answer: 1/3

Example 3hard

In triangle ABC, D is a point on AB and E is a point on AC such that DE is parallel to BC. If AD = x cm, DB = (x - 2) cm, AE = (x + 2) cm, and EC = (x - 1) cm, find the value of x.

A)4 cm
B)5 cm
C)3 cm
D)6 cm

Step-by-step solution

Since DE || BC, by the Basic Proportionality Theorem (BPT), we have AD/DB = AE/EC.
Substitute the given values into the BPT equation.
$x / (x - 2) = (x + 2) / (x - 1)$
Cross-multiply and solve the quadratic equation.
$x(x - 1) = (x - 2)(x + 2) => x² - x = x² - 4 => -x = -4 => x = 4$
Since side lengths must be positive, x - 2 = 4 - 2 = 2 > 0, and x - 1 = 4 - 1 = 3 > 0. So x = 4 is a valid solution.

Answer: 4 cm

Practice questions on Triangles

Q1.easy
Two similar triangles have areas of 121 cm² and 64 cm². If the median of the larger triangle is 11 cm, what is the length of the corresponding median of the smaller triangle?
1. A)6 cm
2. B)7 cm
3. C)8 cm
4. D)9 cm
Show answer
Answer: 8 cm
Hint: Recall the relationship between the ratio of areas of similar triangles and the ratio of their corresponding medians.
Q2.easy
In ΔABC, a line DE is drawn parallel to BC, intersecting AB at D and AC at E. If AD = x cm, DB = (x - 2) cm, AE = (x + 2) cm, and EC = (x - 1) cm, find the value of x.
1. A)4 cm
2. B)5 cm
3. C)6 cm
4. D)7 cm
Show answer
Answer: 4 cm
Hint: Apply the Basic Proportionality Theorem (Thales' Theorem) which states that a line parallel to one side of a triangle divides the other two sides proportionally.
Q3.easy
In ΔABC, D and E are points on sides AB and AC respectively. Which of the following conditions is sufficient to prove that DE || BC?
1. A)AD/AB = AE/AC
2. B)AD/DB = AE/EC
3. C)∠ADE = ∠ABC
4. D)All of the above
Show answer
Answer: All of the above
Hint: Recall the various conditions for the converse of BPT and how similarity implies parallelism.
Q4.medium
In triangle ABC, AD is the median to side BC. If the lengths of the sides are AB = 10 cm, AC = 12 cm, and BC = 16 cm, find the length of the median AD.
1. A)√52 cm
2. B)√58 cm
3. C)√64 cm
4. D)√70 cm
Show answer
Answer: √58 cm
Hint: Drop a perpendicular from A to BC to form right-angled triangles. Use the Pythagorean theorem and properties of medians.
Q5.medium
In triangle ABC, DE is parallel to BC, with D on AB and E on AC. If the area of triangle ADE is 36 cm² and the area of triangle BDE is 24 cm², what is the area of triangle BCE?
1. A)40 cm²
2. B)42 cm²
3. C)48 cm²
4. D)54 cm²
Show answer
Answer: 48 cm²
Hint: Triangles sharing a common altitude have areas proportional to their bases. Use this property and the BPT.
Q6.medium
In triangle ABC, the angle bisector of angle A meets side BC at D. A line through D parallel to AB intersects side AC at E. If the lengths of sides are AC = 21 cm and AB = 14 cm, what is the length of DE?
1. A)4.8 cm
2. B)5.6 cm
3. C)6.3 cm
4. D)7.0 cm
Show answer
Answer: 5.6 cm
Hint: Apply the Angle Bisector Theorem first to find ratios on BC. Then use the property of parallel lines to find similar triangles.
Q7.hard
In a right-angled triangle ABC, right-angled at B, AB = 6 cm and BC = 8 cm. A point D is taken on AC such that BD is perpendicular to AC. What is the length of BD?
1. A)4.2 cm
2. B)4.8 cm
3. C)5.0 cm
4. D)5.2 cm
Show answer
Answer: 4.8 cm
Hint: Consider the area of the triangle and the relationship between the altitude and the hypotenuse in a right-angled triangle.
Q8.hard
Let ABC be an equilateral triangle with side length 's'. P is a point in the interior of the triangle. Perpendiculars PM, PN, PK are drawn from P to sides BC, CA, AB respectively. If PM = 1 unit, PN = 2 units, PK = 3 units, what is the side length 's' of the equilateral triangle?
1. A)2√3 units
2. B)3√3 units
3. C)4√3 units
4. D)5√3 units
Show answer
Answer: 4√3 units
Hint: Divide the equilateral triangle into three smaller triangles with a common vertex P. Relate their areas to the area of the main triangle.
Q9.hard
ABCD is a trapezium with AB || CD. Diagonals AC and BD intersect at O. If AO = (3x-1) cm, OC = (5x-3) cm, BO = (2x+1) cm, and OD = (6x-5) cm, find the value of x.
1. A)1 cm
2. B)2 cm
3. C)3 cm
4. D)4 cm
Show answer
Answer: 1 cm
Hint: Recall the property of similar triangles formed by the diagonals of a trapezium.

These are 9 of the 58 questions available for Triangles. Start practicing above to unlock visual solutions, AI coaching, and the topic leaderboard — completely free.